Question

If a $△ABC$, the bisectors $\angle B$ and $\angle C$ intersect at $O$. Prove that $\angle BOC={90}^o+\frac{1}{2}\angle A$

Answer 1

In $△OBC,\angle OBC=\frac{1}{2}\angle B,\angle OCB=\frac{1}{2}\angle C$

$\angle BOC={180}^{\circ }-\frac{\angle B+\angle C}{2}={180}^{\circ }-\frac{{180}^{\circ }-\angle A}{2}={90}^{\circ }+\frac{1}{2}\angle A$